Question

tan theta/2 + cot theta/2 divided by cot theta/2-tan theta/2 =sec theta​

Answer 1

Given:

tan theta/2 + cot theta/2 divided by cot theta/2-tan theta/2 =sec theta​

To find:

to prove that LHS= RHS

Solution:

please find your solution attached in the image below:

we have proven that LHS = RHS.

Hence, proved.

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Answer 2

GIVEN

Trigonometric equation -

$\frac{ \tan( \frac{θ}{2} ) + \cot( \frac{θ}{2} ) }{ \cot( \frac{θ}{2} ) - \tan(\frac{θ}{2} ) } = \sec( {θ} )$

TO FIND

To prove the given trignonetric equation.

SOLUTION

We can simply solve the above problem as follows -

We are given a trignometric equation.

Let us solve the LHS-

$\frac{ \tan( \frac{θ}{2} ) + \cot( \frac{θ}{2} ) }{ \cot( \frac{θ}{2} ) - \tan(\frac{θ}{2} ) }$

Applying the following formulas in both numerator and denominator -

$\tan(θ) = \frac{ \sin(θ) }{ \cos(θ) }$

And,

$\cot(θ) = \frac{ \cos(θ) }{ \sin(θ) }$

=

$\frac{ \frac{ \sin(θ/2 ) }{ \cos(θ/2 ) } + \frac{ \cos(θ/2 ) }{ \sin(θ/2 ) } }{ \frac{ \cos(θ/2 ) }{ \sin(θ/2 ) } - \frac{ \sin(θ/2 ) }{ \cos(θ/2 ) } }$

Simplifying further -

$\frac{ \frac{ { \sin^{2}(θ/2) } + { \cos^{2}(θ/2) } }{ \sin(θ/2 \cos(θ/2) ) } }{ \sin(θ/2) \cos(θ/2) }$

We know that ,

$\sin^{2}(θ) + { \cos^{2} (θ) } = 1$

Therefore,

$= \frac{1}{ { \cos^{2}(θ/2) - { \sin }^{2}(θ/2) } }$

On simplifying,

$= \frac{1}{ \cos(θ) }$

we know that,

$\frac{1}{ \cos(θ) } = \sec(θ)$

Since, LHS = RHS

Hence,

$\frac{ \tan( \frac{θ}{2} ) + \cot( \frac{θ}{2} ) }{ \cot( \frac{θ}{2} ) - \tan(\frac{θ}{2} ) } = \sec( {θ} )$

Proved.

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